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Chapter 4: Problem 16

If \(46.3 \mathrm{g} \mathrm{PCl}_{3}\) is produced by the reaction $$ 6 \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{P}_{4}(\mathrm{s}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{l}) $$ how many grams each of \(\mathrm{Cl}_{2}\) and \(\mathrm{P}_{4}\) are consumed?

Short Answer

Expert verified
The consumed mass of \(Cl_{2}\) and \(P_{4}\) are 35.84 g and 10.43 g respectively.

Step by step solution

01

Calculate Moles of \(PCl_{3}\)

Firstly, calculate the number of moles of \(PCl_{3}\) produced using the given weight and the molar mass of \(PCl_{3}\). The molar mass of \(PCl_{3}\) is approximately \(30.97+35.45*3 = 137.32 \, g/mol\). So, number of moles = mass/molar mass = \(46.3/137.32 = 0.337 \, mol\)
02

Determine Moles of \(Cl_{2}\) and \(P_{4}\)

Next, use the stoichiometry of the chemical reaction to calculate the moles of \(Cl_{2}\) and \(P_{4}\). From the balanced equation, we can see for every 4 moles of \(PCl_{3}\) produced, 6 moles of \(Cl_{2}\) are consumed and 1 mole of \(P_{4}\) is consumed. Therefore, \(Cl_{2} = 6/4 \times 0.337 = 0.5055 \, mol\) and \(P_{4} = 1/4 \times 0.337 = 0.08425 \, mol\)
03

Calculate Mass of Consumed Reactants

We then convert these mole quantities back to mass using the molar mass of each substance. The molar mass of \(Cl_{2}\) is \(35.45*2 = 70.9 \, g/mol\), and that of \(P_{4}\) is \(30.97*4 = 123.88 \, g/mol\). Therefore, mass of \(Cl_{2} = 0.5055 \times 70.9 = 35.84 \, g\) and mass of \(P_{4} = 0.08425 \times 123.88 = 10.43 \, g\)

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Most popular questions from this chapter

When the equation below is balanced, the correct set of stoichiometric coefficients is (a) \(1,6 \longrightarrow 1,3,4;\) (b) \(1,4 \longrightarrow 1,2,2 ;\) (c) \(2,6 \longrightarrow 2,3,2;\) (d) \(3,8 \longrightarrow 3,4,2\) \(\begin{aligned} ? \mathrm{Cu}(\mathrm{s})+? \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow ? \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+? \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+? \mathrm{NO}(\mathrm{g}) \end{aligned}\)

Nitrogen gas, \(\mathrm{N}_{2}\), can be prepared by passing gaseous ammonia over solid copper(II) oxide, \(\mathrm{CuO}\), at high temperatures. The other products of the reaction are solid copper, \(\mathrm{Cu},\) and water vapor. In a certain experiment, a reaction mixture containing \(18.1 \mathrm{g} \mathrm{NH}_{3}\) and \(90.4 \mathrm{g}\) CuO yields \(6.63 \mathrm{g} \mathrm{N}_{2}\). Calculate the percent yield for this experiment.

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$

Write a chemical equation to represent the complete combustion of malonic acid, a compound with \(34.62 \% \mathrm{C}, 3.88 \% \mathrm{H},\) and \(61.50 \% \mathrm{O},\) by mass.
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